$$\eqalign{ & {\text{Let }}x{\text{ is the number}} \cr & {\text{According to question,}} \cr & 5x = 2{x^2} - 3 \cr & 2{x^2} - 5x - 3 = 0 \cr & 2{x^2} - 6x + x - 3 = 0 \cr & 2x\left( {x - 3} \right) + 1\left( {x - 3} \right) = 0 \cr & \left( {2x + 1} \right)\left( {x - 3} \right) = 0 \cr & x = 3 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & x \times 7 = 333333 \cr} $$
(In answer 5 digit number are given, so we take 6 digit 333333)
$$\eqalign{ & x = \frac{{333333}}{7} \cr & x = 47619 \cr} $$

Sum first n natural numbers
$${\text{ = }}\left[ {\frac{{{\text{n(n + 1)}}}}{2}} \right] = 25 \times 13$$
Sum of first 25 natural numbers
$${\text{ = }}\frac{{25(26)}}{2} = 25 \times 13$$
Now check option
Clearly, we can see 13 is factor
So, '13' is answer

$$\frac{{2}}{{7}}$$ $$\frac{{1}}{{3}}$$ $$\frac{{5}}{{6}}$$ $$\frac{{3}}{{4}}$$ ↓ ↓ ↓ ↓ 0.2857   0.333   0.83   0.75 $$\frac{5}{6}\,{\text{is the greatest value}}$$

$$\frac{{3.157 \times 4126 \times 3.198}}{{63.972 \times 2835.121}}$$     = 0.22
$$\eqalign{ & \Rightarrow {\text{Closest value = 0}}{\text{.2 ( approx}}{\text{.)}} \cr & {\text{ }} \cr} $$

Let number is N
N is divided by the sum of 555 and 445, the quotient is 2 times difference
Quotient = 2 (555 - 445) = 220
Remainder = 30 (given)
N = 1000 × 220 + 30
    = 220030

Let the number is a, b, c
a + b + c = 2
a = $$\frac{1}{2}$$ b.....(i)
c = $$\frac{1}{4}$$ b.....(ii)
$$\frac{a}{b}$$ = $$\frac{1}{2}$$
$$\frac{b}{c}$$ = $$\frac{4}{1}$$
a   :   b   :   c 1 : 2 ↘ ↘ 4 : 1 4 : 8 : 2 a : b : c = 2 : 4 : 1
(2 + 4 + 1) = 7
7 units → 2
1 unit → $$\frac{2}{7}$$
4 units → $$\frac{2}{7}$$ × 4 = $$\frac{8}{7}$$
So, 2nd number is $$\frac{8}{7}$$

Let the numbers are 7x and 9x
According to the question,
$$\eqalign{ & \Rightarrow 7x \times 9x = 1575 \cr & \Rightarrow 63{x^2} = 1575 \cr & \Rightarrow {x^2} = 25 \cr & \Rightarrow x = 5 \cr} $$
Then greater number
= 9x
= 9 × 5
= 45

If first number (729) is completely divide by second number (27) then we divide remainder from second number then remaining remainder will be answer.
$$\frac{729}{27}$$ = 27 times
Now, $$\frac{56}{27}$$ = 2 × 27 + 2 → Remainder

Let the numbers = a, b
According to the question,
a + b = 3 (a - b)
$$\frac{a + b}{a - b}$$   = $$\frac{3}{1}$$
By componendo and dividendo
$$\frac{2a}{2b}$$   = $$\frac{4}{2}$$
$$\frac{a}{b}$$   = $$\frac{2}{1}$$
a : b = 2 : 1

$$\eqalign{ & = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{7} + \frac{1}{{14}} + \frac{1}{{28}} \cr & = \frac{{28 + 14 + 7 + 4 + 2 + 1}}{{28}} \cr & = \frac{{56}}{{28}} \cr & = 2 \cr} $$

$$\eqalign{ & \Rightarrow \left( {{n^3} - n} \right)\left( {n - 2} \right){\text{put n = 3}} \cr & \Rightarrow \left( {{3^3} - 3} \right)\left( {3 - 2} \right) \cr & \Rightarrow \left( {27 - 3} \right) \times 1 \cr & \Rightarrow 24 \cr & {\text{It is divisible by }}24 \cr} $$

$$\eqalign{ & \left( x \right) + \left( {x + 6} \right) + \left( {x + 12} \right) = 63 \cr & 3x + 18 = 63 \cr & 3x = 45 \cr & x = 15 \cr & {\text{So,the largest number is}} \cr & {\text{ = }}15 + 12 \cr & = 27 \cr} $$

$$\eqalign{ & {\text{First number = 1125}} \cr & {\text{Last number = 4950}} \cr & {\text{Number of term }} \cr & {\text{ = }}\frac{{4950 - 1125}}{{225}} + 1 \cr & = \frac{{3825}}{{225}} + 1 \cr & = 17 + 1 \cr & = 18 \cr} $$

$$\eqalign{ & x + \left( {x + 2} \right) + \left( {x + 4} \right) = 54 \cr & 3x + 6 = 54 \cr & 3x = 48 \cr & x = 16 \cr & \cr & {\bf{Alternate: }} \cr & {\text{Middle number = }}\frac{{54}}{3} = 18 \cr & {\text{then number is 16,18,20}} \cr & {\text{Smallest number is 16}} \cr} $$

$$\eqalign{ & {\text{Let the number multiplied by }}x \cr & x \times 0.022 = 66 \cr & x = \frac{{66}}{{22}} \times 1000 = \,3000 \cr} $$

Let total income is 15x (L.C.M. of 3, 5)
$$\eqalign{ & {\text{Spend on food}} \cr & {\text{ = }}\frac{1}{3} \times 15x \cr & = 5x \cr & {\text{Spend on rent}} \cr & {\text{ = }}\frac{2}{5} \times 15x \cr & = 6x \cr & {\text{Spend on clothes }} \cr & {\text{ = }}\frac{1}{5} \times 15x \cr & = 3x \cr & {\text{Income left }} \cr & = 15x - \left( {5 + 6 + 3} \right)x \cr & = 15x - 14x = x \cr & x = 400 \cr & 15x = 6000 \cr & {\text{Income = 6000 }} \cr} $$

Dividend = divisor × quotient + remainder
$$\eqalign{ & {\text{First number }} \cr & \Rightarrow {\text{ }}\left( {17 \times {\text{n}}} \right) + 13 \cr & {\text{Let n = 1}} \cr & \Rightarrow (17 \times 1) + 13 \cr & \Rightarrow 30 \cr & {\text{second number }} \cr & \Rightarrow {\text{ }}\left( {17 \times {\text{n}}} \right) + 11 \cr & \Rightarrow (17 \times 1) + 11 \cr & \Rightarrow 28 \cr & {\text{According to question}} \cr & = \frac{{30 + 28}}{{17}} \cr & = \frac{{58}}{{17}} \cr & \Rightarrow {\text{remainder = 7}} \cr} $$

Alternate :
Divisor = 1st remainder + 2nd remainder - 3rd remainder
$$\eqalign{ & 17 = {\text{ }}13 + 11 - {\text{3rd remainder}} \cr & {\text{3rd remainder}} = 24 - 17 = 7 \cr} $$

According to the question,
$$ \Rightarrow \left( {2 - \frac{1}{3}} \right)\left( {2 - \frac{3}{5}} \right)\left( {2 - \frac{5}{7}} \right).....$$       $$\left( {2 - \frac{{997}}{{999}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{5}{3} \times \frac{7}{5} \times \frac{9}{7}.....\frac{{1001}}{{999}} \cr & \Rightarrow \frac{1}{3} \times 1001 \cr & \Rightarrow \frac{{1001}}{3} \cr} $$

Shortcut Method :
Divisor = 1st remainder + 2nd remainder - 3rd remainder
$$\eqalign{ & = 3 + 4 - 2{\text{ }} \cr & = 7 - 2 \cr & = 5 \cr} $$